Question

# Armer Company is accumulating data to use in preparing its annual profit plan for the coming...

Armer Company is accumulating data to use in preparing its annual profit plan for the coming year. The cost behavior pattern of the maintenance costs must be determined. The accounting staff has suggested the use of linear regression to derive an equation for maintenance hours and costs. Data regarding the maintenance hours and costs for the last year and the results of the regression analysis follow:

 Month Maintenance Cost Machine Hours Jan. \$ 4,950 780 Feb. 3,750 620 Mar. 4,350 700 Apr. 3,570 600 May 5,100 800 June 3,710 610 July 3,780 620 Aug. 5,220 880 Sept. 5,010 790 Oct. 4,800 770 Nov. 4,050 650 Dec. 3,910 640 Sum \$ 52,200 8,460 Average \$ 4,350 \$ 705 Average cost per hour \$ 6.00 a (intercept) \$ −238.58 b (coefficient) 6.5086 Standard error of the estimate 114.594 R-squared 0.96949 t-value for b 17.8260

At 400 hours of activity, Armer management can be approximately two-thirds confident that the maintenance costs will be in the range of __________

Multiple Choice

• \$2,105.33 to \$2,624.39.

• \$2,250.27 to \$2,479.45.

• \$2,221.14 to \$2,508.70.

• None of these answer choices are correct.

• \$2,106.28 to \$2,623.41.

SE = \$114.594

Predicted cost at 400 hours = -\$238.58 +400*6.5086

=\$2,364.86

Confidence range = \$2,364.86 +(or)-\$114.594

\$2,250.27 to \$ 2,479.45

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